We could have used these relationships to determine A1,
A2, and A3. But A1 and A3
were easily found using the "cover-up"
method. The top relationship tells us that A2=-0.25, so

and

(where, again, it is implicit that f(t)=0 when t<0).

Many texts use a method based upon differentiation of the fraction when there
are repeated roots. The technique involves differentiation of ratios of
polynomials which is prone to errors.
Details are
here if you are interested.

Example: Complex Conjugate Roots (Method 1)

Solution:If we use complex roots, we can expand the fraction as we did before.
This is not typically the way you want to proceed if you are working by
hand, but may be easier for computer solutions (where complex numbers are
handled as easily as real numbers). To perform the expansion, continue
as before.

where

Note that A2 and A3 must be complex conjugates of each
other since they are equivalent except for the sign on the imaginary part.
Performing the required calculations:

Example: Complex Conjugate Roots (Method
2)

Solution:Another way to expand the fraction without resorting to complex numbers
is to perform the expansion as follows.

Note that the numerator of the second term is no longer
a constant, but is instead a first order polynomial. From above (or using the
cover-up method) we know that A=-0.2. We can find the quantities B and
C from cross-multiplication.

If we equate like powers of "s" we get

order of coefficient

left sidecoefficient

right sidecoefficient

2nd (s2)

0

A+B

1st (s1)

1

4A+5B+C

0th (s0)

3

5A+5C

Since we already know that
A=-0.2, the first expression (0=A+B) tells us that B=0.2, and
the last expression (3=5A+5C) tells us that C=0.8. We can use
the middle expression (1=4A+5B+C) to check our calculations.
Finally, we get

The two previous examples have demonstrated two techniques for
performing a
partial fraction expansion of a term with complex roots.
The first technique was a simple extension of the rule for
dealing with distinct real roots. It is conceptually
simple, but can be difficult when working by hand because of the
need for using complex numbers; it is easily done by computer.
The second technique is easy to do by hand, but is conceptually
a bit more difficult. It is easy to show that the two
resulting partial fraction representations are equivalent to each other.
Let's first examine the result from Method 1 (using two techniques).

We now repeat this calculation, but in the process we develop a general
technique (that proves to be useful when using MATLAB to help with the
partial fraction expansion. We know that F(s) can be represented as a
partial fraction expansion as shown below:

We can now find the inverse
transform of the complex conjugate terms by treating them as
simple first order terms (with complex roots).

In this expression M=2K. The frequency (ω)
and decay coefficient (σ) are determined from the root of the denominator of A2 (in this
case the root of the term is at s=-2+j; this is where the term is equal to
zero). The frequency is the
imaginary part of the root (in this case, ω=1), and the decay coefficient is the real part of the root (in this case,
σ=-2).

Using the cover-up method (or, more likely, a
computer program) we get

and

This yields

It is easy to show that the
final result is equivalent to that previously found, i.e.,

While this method is somewhat
difficult to do by hand, it is very convenient to do by
computer. This is the approach used on the page that shows
MATLAB techniques.

Finally we present Method 2, a technique that is easier to work with
when solving problems for hand (for homework or on exams) but is less useful
when using MATLAB.

Thus it has been shown that the two
methods yield the same result. Use Method 1 with MATLAB and use Method 2
when solving problems with pencil and paper.

Example - Combining multiple expansion methods

Find the inverse Laplace Transform of

Solution:The fraction shown has a second order term in the denominator that
cannot be reduced to first order real terms.
As discussed in the page describing
partial
fraction expansion, we'll use two techniques. The
first technique involves expanding the fraction while retaining the second order
term with complex roots in the denominator. The second technique
entails "Completing the Square."

We repeat the previous example, but use a brute force technique.
You will see that this is harder to do when solving a problem manually,
but is the technique used by MATLAB. It is important to be able to
interpret the MATLAB solution.

Find the inverse Laplace Transform of

Solution:We can express this as four terms, including two complex terms (with A3=A4*)

Cross-multiplying we get (using the fact that
(s+1-2j)(s+1+2j)=(s2+2s+5))

We will use the notation derived above (Method 1 - a more general technique).
The root of the denominator of the A3 term in the partial
fraction expansion is at s=-1+2j (i.e., the denominator goes to 0 when
s=-1+2j), the magnitude of A3 is √2, and the angle of A3
is 225°. So, M=2√2, φ=225°,
ω=2, and σ=-1. Solving for f(t) we get

This expression is equivalent to the one obtained
in the previous example.

The last case we will consider is that of exponentials in the numerator of
the function.

Example: Exponentials in the numerator

Find the inverse Laplace Transform of the function F(s).

Solution:The exponential terms indicate a time delay
(see the time delay property).
The first thing we need to do is collect terms that have the same time
delay.

We now perform a partial fraction expansion for each
time delay term (in this case we only need to perform the expansion for the
term with the 1.5 second
delay), but in general you must do a complete expansion for each term.

Now we can do the inverse Laplace Transform of each term
(with the appropriate time delays)